To determine $x$ in the vector $q = (152, 46, x)$ that belongs to the span of $a = (-4, 3, -10)$ and $b = (-36, -5, -13)$, we express $q$ as a linear combination of $a$ and $b$:

$q = \alpha a + \beta b$

where $\alpha$ and $\beta$ are scalars, and $q = (152, 46, x)$. This means we need to solve for $\alpha$ and $\beta$ in the system of equations:

$(152, 46, x) = \alpha(-4, 3, -10) + \beta(-36, -5, -13)$

This yields the following system of three equations:

1. $152 = \alpha(-4) + \beta(-36)$
2. $46 = \alpha(3) + \beta(-5)$
3. $x = \alpha(-10) + \beta(-13)$

Now we can solve these equations.

Step 1: Solve for $\alpha$ and $\beta$ from the first two equations.

From the first equation:

$152 = -4\alpha - 36\beta$ $4\alpha + 36\beta = -152$ Divide by 4: $\alpha + 9\beta = -38 \quad \text{(Equation 1)}$

From the second equation:

$46 = 3\alpha - 5\beta$ $3\alpha - 5\beta = 46 \quad \text{(Equation 2)}$

Step 2: Solve the system of equations.

Substituting Equation 1 into Equation 2:

From Equation 1: $\alpha = -38 - 9\beta$.

Substitute this into Equation 2:

$3(-38 - 9\beta) - 5\beta = 46$ $-114 - 27\beta - 5\beta = 46$ $-114 - 32\beta = 46$ $-32\beta = 46 + 114$ $-32\beta = 160$ $\beta = -5$

Solve for $\alpha$:

Substitute $\beta = -5$ back into Equation 1:

$\alpha + 9(-5) = -38$ $\alpha - 45 = -38$ $\alpha = 7$

Step 3: Solve for $x$.

Now substitute $\alpha = 7$ and $\beta = -5$ into the third equation:

$x = \alpha(-10) + \beta(-13)$ $x = 7(-10) + (-5)(-13)$ $x = -70 + 65$ $x = -5$

Thus, the value of $x$ is $\boxed{-5}$.

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Would you like a more detailed explanation of any step? Or do you have further questions?

Related questions:

1. How can I find the span of two vectors in $\mathbb{R}^3$?
2. What are the properties of linear combinations in vector spaces?
3. How do I solve systems of linear equations using substitution or elimination?
4. What is the concept of vector projection and how does it relate to spans?
5. Can vectors in $\mathbb{R}^3$ ever be linearly independent or dependent?

Tip: When solving systems of linear equations, remember that you can always use matrix methods (like Gaussian elimination) to find solutions efficiently!